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\begin{center}\framebox{\parbox{\boxlength}{\bf
CS 681: Computational Number Theory and Algebra \hfill Lecture 9
\\\\
Polynomial factorization over Finite Fields
\\\\
Lecturer: Manindra Agrawal \hfill Scribe: Sudeepa Roy
%\\
\begin{flushright}
%date
August 19, 2005
\end{flushright}
}}\end{center} \vspace{5mm}

\section{Introduction }
In the last lecture we studied the tool automorphism over finite rings. In this lecture we will discuss how to use automorphism to factorize a polynomial over finite fields.
\\Let $f(x)$ be a polynomial of degree $d$ over field $F_q$.
\begin{definition}
$f$ is \textbf{square free} if $g^2$ does not divide for any $g$.
\end{definition}


\section{Factorization algorithms for different types of polynomials }
\subsection{Case I : $f$ is not square free}
In this case $g^2~ |~ f$ for some $g$.
\\Let $\frac{df}{dx} = f'$.
\\Then $g~ |~ gcd(f,f')$.
\\This produces a factor of $f$.

\subsection{Case II : $f$ is square free}
Let $f~ =~ f_1  f_2  \cdots  f_k$
\\where each $f_i$ is irreducible and let $\deg f_i~ =~ d_i$ with
\begin{center}
$d_1 \leq d_2 \leq \cdots \leq d_k$
\end{center}
Let 
%$\left[ abc \right]$s\\
$R = F_q[X] / (f(X))$
\\ $= \oplus_{i=1}^{k} F_q[X] / (f_i(X))$ 
\\$[$ by Chinese Remaindering, as all the $f_i$ s are distinct and irreducible, so are prime to each other $]$.
\\\\Let 

$\psi(y)~ =~ y^q$
\begin{observation}
$\psi$ is an automorphism of $F_q[X]/(f_i(X))$ and $\psi^j$ = \textit{id} in $F_q[X]/(f_i(X))$ iff $j~ =~ d_i$.
\end{observation}

\subsubsection{Case II.1 : There is an i such that $d_i > d_1$}
Then $\psi^{d_1}$ is trivial in $F_q[X] / (f_1(X))$ but not in $F_q[X] / (f_i(X))$.
\\In other words,

$\psi^{d_1}(X)~ -~ X~ =~ 0$ in $F_q[X] / (f_1(X))$ but not in $F_q[X] / (f_i(X))$
\\$\Rightarrow f_1(x)~|~ \psi^{d_1}(x)~ -~ x$ but \textbf{\textit{not}} $f_i(x)~|~ \psi^{d_1}(x)~ -~ x$
\\\\{\large\textbf{Algorithm}}
%\normalsize
\\\\$for$ $i~=~1$ $to$ $d-1$ $do$

compute $gcd(\psi^i(x)~ -~x,~f(x))$
\\\\{\large\textbf{Time Complexity}
\begin{observation}
$gcd(\psi^i(x)~ -~x,~f(x))~ =~ gcd((\psi^i(x)~ -~x) \mod f(x),~f(x))$
\end{observation}
Hence in each step of the algorithm we will perform $=~ gcd(x^{q^i} \mod f(x) -~ x,~f(x))$ so that the degree of both the terms are bounded above by $\deg f(x)~ =~ d$.
\\\\To compute $x^{q^i}$ we will follow repeated squaring method, where we will compute the sequence $x,x^2,x^4,\cdots,x^{2^j}$ $[$ each modulo f $]$ unless $2^j > q^i$.
\\Here no. of squaring required = $\log{q^i}~ =~ i\log q~ \leq~ d\log q$ as $i~ \leq~ d$.
\\\\Using FFT, complexity of polynomial multiplication = complexity of polynomial division = $O(d \log d)$ where degree of each polynomial is bounded by $d$. So at each step of the above sequence computation, multiplication and taking modulo $f$ needs $O(d \log d)$ operations.
As each element of the field $F_q$ is $\log q$ bits long, so complexity of multiplication of coefficients of $f$ using FFT is $O(\log q~ \log\log q~ \log\log\log q$), or ignoring sublogarithmic factors $\tilde{O}(\log q)$.
\\\\ Hence time complexity to compute $x^{q^i}$
\\   = $\tilde{O}(d \log q.~ d \log d.~ \log q)$\\
   = $\tilde{O}(d^2~ (\log q)^2~ \log d)$\\
	 = $\tilde{O}(d^2 \log^2 q)$ $[$ ignoring $\log d$ factor $]$
\\\\ To compute $gcd(\psi^i(x)~ -~x,~f(x))$
\\   = $\tilde{O}(d^3 \log^2 q)$ $[$ as we may have to iterate at most $d$ times to get the gcd $]$.
\\\\Hence to iterate the procedure $d-1$ times, time complexity of the algorithm = $\tilde{O}(d^4\log^2q)$.
\\$[$Using more intelligent gcd algorithm the time complexity can be reduced by a factor of $d$ $]$.
\subsubsection{Case II.2 : $d_1 = d_2 = \cdots = d_k = \frac{d}{k}$}
In this case, $gcd(\psi^i(x)~ -~x,~f(x))~ =~ 1$ for $i~ < \frac{d}{k}$ and $gcd(\psi^{\frac{d}{k}}(x)~ -~x,~f(x))~ =~ f(x)$.
\\Hence we can obtain no. of factors of the polynomial $f$, if we note down the point $i=t$ such that the value of the gcd changes from $1$, then $\frac{d}{t}~ =~ k$ = no. of factors of $f$.
\\\\The first step will be to reduce the problem to finding roots $[$ finding roots is equivalent to find the linear factors of $f$, so it is no harder than factorization problem $]$.
\\\\$R = \oplus_{i=1}^{k} F_q[X] / (f_i(X))$ $[$ by Chinese Remaindering, as all the $f_i$s are distinct and irreducible $]$
\\Let $S = \left\{ e(X)~ |~ e(X)\in R ~\&~ \psi(e(X))~ =~ e(X)\right\}$
\\Each $e(X)$ can be viewed as a k-tuple.
\begin{observation}
Each component of $e(X)\in S$ represented as a k-tuple $\in F_q$.
\end{observation}
Hence,
\\ $|S|~=~q^k~ >~ q~ =~ |F_q|$ if $k~ >~ 1$.
\\$[$We have $k$ tuples and $q$ elements in each tuple, and by Chinese Remaindering Theorem all are distinct elements of $R$ $]$
\\$\Rightarrow$ There is an $e(X)~ \in S~ -~ F_q$
\\\\$[$ To be continued in the next lecture $]$.
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